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Question Number 163902 by HongKing last updated on 11/Jan/22

$$\sqrt{\mathrm{x}{!}^{\mathbf{x}!}}+2^{\mathbf{x}!}=\mathrm{x}{!}^3+10\mathrm{x}!+4$$$$\mathrm{find}:\mathbf{x}=?$$

Answered by MJS_new last updated on 11/Jan/22

$$x!\mathrm{is}\mathrm{defined}\mathrm{for}x\in \mathbb{N}\Rightarrow \mathrm{try}\mathrm{the}\mathrm{first}\mathrm{few}...$$$$\Rightarrow x=3$$

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